The lower bound for the volume of a three-dimensional convex polytope
In this paper, we provide a lower bound for the volume of a three-dimensional smooth integral convex polytope having interior lattice points. Our formula has a quite simple form compared with preliminary results. Therefore, we can easily utilize it for other beneficial purposes. Firstly, as an immed...
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Інститут прикладної математики і механіки НАН України
2015
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| Назва видання: | Algebra and Discrete Mathematics |
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| Цитувати: | The lower bound for the volume of a three-dimensional convex polytope / R. Kawaguchi // Algebra and Discrete Mathematics. — 2015. — Vol. 20, № 2. — С. 263-285. — Бібліогр.: 17 назв. — англ. |
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irk-123456789-1551392019-06-17T01:26:49Z The lower bound for the volume of a three-dimensional convex polytope Kawaguchi, R. In this paper, we provide a lower bound for the volume of a three-dimensional smooth integral convex polytope having interior lattice points. Our formula has a quite simple form compared with preliminary results. Therefore, we can easily utilize it for other beneficial purposes. Firstly, as an immediate consequence of our lower bound, we obtain a characterization of toric Fano threefold. Besides, we compute the sectional genus of a three-dimensional polarized toric variety, and classify toric Castelnuovo varieties. 2015 Article The lower bound for the volume of a three-dimensional convex polytope / R. Kawaguchi // Algebra and Discrete Mathematics. — 2015. — Vol. 20, № 2. — С. 263-285. — Бібліогр.: 17 назв. — англ. 1726-3255 2010 MSC:Primary 52B20; Secondary 14C20, 14J30, 14M25. http://dspace.nbuv.gov.ua/handle/123456789/155139 en Algebra and Discrete Mathematics Інститут прикладної математики і механіки НАН України |
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Digital Library of Periodicals of National Academy of Sciences of Ukraine |
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English |
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In this paper, we provide a lower bound for the volume of a three-dimensional smooth integral convex polytope having interior lattice points. Our formula has a quite simple form compared with preliminary results. Therefore, we can easily utilize it for other beneficial purposes. Firstly, as an immediate consequence of our lower bound, we obtain a characterization of toric Fano threefold. Besides, we compute the sectional genus of a three-dimensional polarized toric variety, and classify toric Castelnuovo varieties. |
| format |
Article |
| author |
Kawaguchi, R. |
| spellingShingle |
Kawaguchi, R. The lower bound for the volume of a three-dimensional convex polytope Algebra and Discrete Mathematics |
| author_facet |
Kawaguchi, R. |
| author_sort |
Kawaguchi, R. |
| title |
The lower bound for the volume of a three-dimensional convex polytope |
| title_short |
The lower bound for the volume of a three-dimensional convex polytope |
| title_full |
The lower bound for the volume of a three-dimensional convex polytope |
| title_fullStr |
The lower bound for the volume of a three-dimensional convex polytope |
| title_full_unstemmed |
The lower bound for the volume of a three-dimensional convex polytope |
| title_sort |
lower bound for the volume of a three-dimensional convex polytope |
| publisher |
Інститут прикладної математики і механіки НАН України |
| publishDate |
2015 |
| url |
http://dspace.nbuv.gov.ua/handle/123456789/155139 |
| citation_txt |
The lower bound for the volume of a three-dimensional convex polytope / R. Kawaguchi // Algebra and Discrete Mathematics. — 2015. — Vol. 20, № 2. — С. 263-285. — Бібліогр.: 17 назв. — англ. |
| series |
Algebra and Discrete Mathematics |
| work_keys_str_mv |
AT kawaguchir thelowerboundforthevolumeofathreedimensionalconvexpolytope AT kawaguchir lowerboundforthevolumeofathreedimensionalconvexpolytope |
| first_indexed |
2025-07-14T07:13:57Z |
| last_indexed |
2025-07-14T07:13:57Z |
| _version_ |
1837605572294213632 |
| fulltext |
Algebra and Discrete Mathematics RESEARCH ARTICLE
Volume 20 (2015). Number 2, pp. 263–285
© Journal “Algebra and Discrete Mathematics”
The lower bound for the volume
of a three-dimensional convex polytope
Ryo Kawaguchi
Communicated by V. A. Artamonov
Abstract. In this paper, we provide a lower bound for
the volume of a three-dimensional smooth integral convex polytope
having interior lattice points. Since our formula has a quite simple
form compared with preliminary results, we can easily utilize it for
other beneficial purposes. As an immediate consequence of our lower
bound, we obtain a characterization of toric Fano threefold. Besides,
we compute the sectional genus of a three-dimensional polarized
toric variety, and classify toric Castelnuovo varieties.
1. Introduction
Points in Z
n are called lattice points of Rn, and a polytope is said to be
integral if all its vertices are lattice points. For an integral polytope P , we
denote by vol(P) the volume of P and by ∂P the boundary of P , and put
Int(P) = P \ ∂P. Besides, we define l(S) = ♯(S ∩Z
n) for a subset S ⊂ R
n.
In the study of integral polytopes, one of the most significant problem
is to compute their volume by using the information of the number of
lattice points in them. The following classical theorem gave a clue to the
solution of this issue.
Theorem 1.1 (cf. [14]). Let P be an integral polygon which is homeo-
morphic to a closed circle. Then its volume is computed by 2vol(P) =
l(P) + l(Int(P)) − 2.
2010 MSC: Primary 52B20; Secondary 14C20, 14J30, 14M25.
Key words and phrases: lattice polytopes, polarized varieties, toric varieties,
sectional genus.
264 The lower bound for the volume of a polytope
In the case where an integral polygon is not homeomorphic to a closed
circle, Reeve generalized the above Pick’s result by employing the Euler
characteristics χ(P) of the polygon and χ(∂P) of its boundary.
Theorem 1.2 (cf. [15]). Let P be an n-dimensional integral polygon.
(i) If n = 2, then 2vol(P) = l(P) + l(Int(P)) − 2χ(P) + χ(∂P).
(ii) If n = 3, then
2k(k2 − 1)vol(P) = l(kP) + l(Int(kP)) − k(l(P)
+ l(Int(P))) + (k − 1)(2χ(P) − χ(∂P)),
l(∂(kP)) = k2l(∂P) − 2(k2 − 1)
for any positive integer k, where kP denotes the dilated polytope
{kx | x ∈ P}.
Moreover, Macdonald established the general formula to compute the
volume of an integral polytope of arbitrary dimension in [10]. Concretely,
for an n-dimensional integral polytope P,
(n − 1)n! vol(P) =
n−1
∑
k=1
(−1)n−k−1
(
n − 1
k
)
(l(kP)
+ l(Int(kP))) + (−1)n−1(2χ(P) − χ(∂P)) (1)
and
n! vol(P) =
n
∑
k=1
(−1)n−k
(
n
k
)
l(kP) + (−1)nχ(P). (2)
In addition, some other interesting formulae have been provided by
Kołodziejczyk and Reay in [7–9]. One can in principle compute the volume
of a polytope by using these results. In fact, however, it is not easy to carry
it out. This difficulty comes from the intricate behavior of the number of
lattice points in a dilated polytope kP. Therefore, from the application
standpoint, it is desirable to find a more simple formula even if not as
strong as (1) and (2). Specifically, in this paper, we will give a lower bound
for the volume of a three-dimensional integral convex polytope (Theorem
1.4). Although this result gives only an inequality, it is of wide application
because of its simplicity. First, Corollary 1.6 provides a characterization of
toric Fano threefold. Furthermore, we will apply this corollary to compute
the sectional genus of a three-dimensional polarized toric variety and
R. Kawaguchi 265
classify so-called Castelnuovo varieties in Section 3 (Theorem 3.8). Before
describing our main result, we need to define the smoothness of a polytope.
A polytope is said to be convex if it is a convex hull of a finite number of
points in R
n.
Definition 1.3. Let P be an n-dimensional integral convex polytope in R
n,
and P be a vertex of P. Define R>0(P−P )={a(Q−P )∈R
n |Q∈P, a>0}.
If there exists a Z-basis {m1, . . . , mn} of Zn such that
R>0(P − P ) = R>0m1 + · · · + R>0mn,
the vertex P is said to be smooth. We say P is smooth if all its vertices
are smooth. An m-dimensional (m < n) integral convex polytope P ′ in
R
n is said to be smooth if it is smooth with respect to R
m which is the
smallest affine subspace of Rn including P ′.
Theorem 1.4. Let P be a three-dimensional smooth integral convex
polytope having at least one interior lattice point. Then 3vol(P) > l(P) +
l(Int(P)) − 4, and equality holds if and only if P is a polytope associated
to the anti-canonical bundle on a toric Fano threefold.
Toric Fano threefolds have been already classified into eighteen types in
[2] and [17], independently. Namely, there exist eighteen polytopes (see, e.g.,
(6) in Section 2) whose volume achieves the lower bound in Theorem 1.4.
We remark that the conditions of smoothness and l(Int(P)) > 1 are
essential for the above theorem. Indeed, if we remove these conditions,
we can easily find counterexamples as follows.
Example 1.5. For a subset S of R3, we denote by Conv(S) the convex
hull of S.
(i) For a nonsmooth integral convex polytope
P1 = Conv({(0, 0, ±1), (2, 1, ±1), (1, 2, ±1), (1, 1, 2)}),
we have 3vol(P1) = 21/2 < l(P1) + l(Int(P1)) − 4 = 11.
(ii) For a unit cube P2, we have 3vol(P2)=3 < l(P2)+l(Int(P2))−4=4.
On the other hand, as is well known, the theory of polytopes is
closely related to the toric geometry. For an ample line bundle L on
an n-dimensional compact toric variety X, there exists an associated
n-dimensional integral convex polytope �L from which we can read off
many invariants of L. A computation of the dimension of a cohomology
266 The lower bound for the volume of a polytope
group can be reduced to counting lattice points in the polytope. For
example, we have h0(X, L) = l(�L) and h0(X, L + KX) = l(Int(�L)).
The degree of L can be computed as Ln = n!vol(�L). These relations,
for example, tell us that Pick’s formula coincides with the Riemann-Roch
theorem on a surface. Indeed, if X is two-dimensional, the equalities
χ(OX) = 1 and h0(X, KX) = h1(X, KX) = 0 hold. Besides, the general
theory of toric varieties gives that h1(X, L) = h2(X, L) = 0 if |L| has no
base points. Therefore, we can deform Theorem 1.1 as
h0(X, L) = L2 − h0(X, L + KX) + 2
χ(OX(L)) = L2 − h0(X, KL) + 2 =
1
2
L.(L − KX) + χ(OX).
In this manner, properties of polytopes and that of line bundles on a
toric variety can be translated each other.
Using the terminology of the algebraic geometry, we can interpret
Theorem 1.4 as a theorem about line bundles.
Corollary 1.6. Let L be an ample line bundle on a three-dimensional
smooth compact toric variety X. If h0(X, L + KX) > 1, then L3 >
2(h0(X, L) + h0(X, L + KX) − 4) holds, and equality holds if and only if
X is a toric Fano threefold and L = −KX .
2. Proof of the main theorem
First of all, we need to introduce several notations. We denote by
Hf(x,y,z) the plane in R
3 defined by an equation f(x, y, z) = 0. For a
lattice point P and a polygon F included in a plane Hf(x,y,z), we denote
by h(F, P ) the lattice distance. In concrete terms, we define h(F, P ) = |n|,
where n is an integer such that Hf(x,y,z)−n passes through P .
Henceforth, the notation P always denotes a three-dimensional smooth
integral convex polytope having interior lattice points. In addition, we
denote by V (P) the set of vertices of P, and E(P) the set of points on
edges of P. Note that vol(P), l(P) and l(Int(P)) do not change even if
we perform an affine linear transformation (i.e., a composition of parallel
displacements and linear transformations by unimodular matrices).
Lemma 2.1. If we place P in R
3
z>0 so that P has a face on Hz, then
P ∩ Hz−1 is an integral convex polygon.
Proof. We can assume that the origin O is a vertex of P . Then O has three
adjacent lattice points (a1, b1, 0), (a2, b2, 0), (a3, b3, c3) ∈ E(P). Since the
R. Kawaguchi 267
vertex O is smooth, we have c3 = 1 by the equality
det
a1 a2 a3
b1 b2 b3
0 0 c3
= (a1b2 − a2b1)c3 = ±1.
By similar arguments, we see that any edge of P which is extending
from a vertex on Hz but not lying on Hz has a lattice point on Hz−1.
The assertion follows from this fact.
Lemma 2.2. There exists a plane Hf(x,y,z) such that the section T =
P ∩ Hf(x,y,z) is an integral convex polygon having a smooth vertex and at
least one interior lattice point.
Proof. Without loss of generality, we can assume that O, (1, 0, 0), (0, 1, 0)
and (0, 0, 1) are contained in E(P). Put P1 = (0, 1, 1), P2 = (1, 0, 1) and
P3 = (1, 1, 0). In the case where P1 /∈ P, the integral convex polygon
P ∩ Hx must be a unit triangle by the smoothness of P. Similarly, if P2
(resp. P3) is not contained in P, then P ∩ Hy (resp. P ∩ Hz) becomes
a unit triangle. Since l(Int(P)) > 1 and every vertex of P has just
three edges, it is required that at least two points of P1, P2 and P3 are
contained in P. Hence we can assume (after permuting the coordinates,
if necessary) that P1, P2 ∈ P. It is sufficient to consider the case where
(1, 1, 1) /∈ Int(P), because if (1, 1, 1) ∈ Int(P), we can finish the proof by
putting f(x, y, z) = z − 1.
Let (x0, y0, z0) be an interior lattice point of P. Suppose that P3 ∈ P.
Since (1, 1, 1) is not contained in Int(P), there exist four integers α, β, γ
and δ such that P ⊂ {(x, y, z) | αx+βy+γz+δ > 0} and α+β+γ+δ 6 0.
By the conditions P1, P2, P3 ∈ P, we have
β + γ + δ > 0, α + γ + δ > 0, α + β + δ > 0,
which imply that α, β, γ 6 0. Then we obtain a contradiction
αx0 + βy0 + γz0 + δ > 0
δ > −αx0 − βy0 − γz0 > −α − β − γ.
Thus we see that P3 is not contained in P. In this case, the face
P ∩Hz is a unit triangle, and the vertex (1, 0, 0) has three adjacent lattice
points O, (0, 1, 0) and (a, 0, c) in E(P). One can check c = 1 by the
smoothness of the vertex (1, 0, 0) in a similar way to that in the proof
of Lemma 2.1. Thus P has a face included in the plane Hx+y+(1−a)z−1
268 The lower bound for the volume of a polytope
containing three points (1, 0, 0), (0, 1, 0) and (a, 0, 1). Then we obtain
a > 2 by the inequality
0 > x0 + y0 + (1 − a)z0 − 1 > (1 − a)z0.
It follows that (2, 0, 1) is contained in P . Similarly, one can verify that
(0, 2, 1) is contained in P. By the assumption (1, 1, 1) /∈ Int(P), the section
P ∩ Hz−1 must be a triangle Conv({(0, 0, 1), (2, 0, 1), (0, 2, 1)}), which
implies a = 2. Next we focus on the point (1, 1, 2). Suppose that (1, 1, 2)
is not contained in Int(P). Then there exist four integers ε, ζ, η and θ
such that P ⊂ {(x, y, z) | εx + ζy + ηz + θ > 0} and ε + ζ + 2η + θ 6 0.
Since (0, 0, 1), (2, 0, 1), (0, 2, 1) ∈ P, we have
η + θ > 0, 2ε + η + θ > 0, 2ζ + η + θ > 0,
which imply that ε + η + θ > 0, ζ + η + θ > 0 and ε + ζ + η + θ > 0.
It follows that η 6 −ε − ζ − η − θ 6 0, ε + η 6 −ζ − η − θ 6 0 and
ζ + η 6 −ε − η − θ 6 0. Since P has a face included in Hx+y−z−1, the
inequality x0 + y0 − z0 − 1 < 0 holds, which implies a contradiction
εx0 + ζy0 + ηz0 + θ > 0
θ > −εx0 − ζy0 − ηz0 > −(ε + η)x0 − (ζ + η)y0
> −ε − ζ − 2η.
Therefore, we can conclude that (1, 1, 2) is contained in Int(P), and
P ∩ Hx−1 is the desired section.
Let Q be a three-dimensional integral convex polytope, and Q be a
vertex of Q. We define
µ(Q) = vol(Q) −
l(Q) + l(Int(Q)) − 4
3
and QQ = Conv((Q ∩ Z
3) \ {Q}), and denote by Q
Q
the set of points in
QQ which are visible from Q, that is,
Q
Q
= {P ∈ QQ | (the segment PQ) ∩ QQ = {P}}.
Although Q
Q
is not a convex polytope but a set consisting of some
faces of QQ, we formally define V (Q
Q
) = V (QQ) ∩ Q
Q
, ∂Q
Q
= ∂Q ∩ Q
Q
and Int(Q
Q
) = Q
Q
\ ∂Q
Q
. The following proposition tells us the precise
difference between Q and QQ, which is a central tool in the induction
step in the proof of Theorem 1.4.
R. Kawaguchi 269
Proposition 2.3. Let Q be a vertex of a three-dimensional integral convex
polytope Q, and F1, . . . , Fk be faces of QQ such that Q
Q
=
⋃k
j=1 Fj. If we
put aj = l(Fj) and bj = l(Int(Fj)), then
µ(Q) = µ(QQ) +
1
6
l(∂Q
Q
) −
2
3
+
1
6
k
∑
j=1
(h(Fj , Q) − 1)(aj + bj − 2).
Proof. For simplicity, we put hj = h(Fj , Q). Since vol(Fj) = (aj +bj −2)/2
by Theorem 1.1, we have
µ(Q) − µ(QQ) =
1
6
k
∑
j=1
hj(aj + bj − 2) −
1
3
(l(Int(Q)) − l(Int(QQ)) + 1)
=
1
6
k
∑
j=1
(hj − 1)(aj + bj − 2) +
1
6
k
∑
j=1
(aj + bj)
−
1
3
(l(Int(Q
Q
)) + k + 1).
To estimate the right-hand value, let us compute
∑k
j=1(aj + bj), which
means counting lattice points in Q
Q
(with several duplications). Indeed,
we can write
k
∑
j=1
(aj + bj) =
∑
P ∈Q
Q
∩Z3
c(P ), (3)
where c(P ) denotes the number of times P is counted in the left-hand
side of (3). It is clear that c(P ) = 1 for P ∈ ∂Q
Q
\ V (Q
Q
) since there
exists a unique face containing P in this case. Let us check c(P ) = 2
for a point P in Int(Q
Q
) \ V (Q
Q
). This is clear if P is not on any edge
of QQ. While if P is on some edge of QQ, then there exist two faces
of Q
Q
containing P . Hence, in this case, P is counted two times in the
left-hand side of (3). We next consider points in V (Q
Q
). Let P be a point
in V (Q
Q
) ∩ Int(Q
Q
) having s edges. Since s faces contain P , P is counted
s times in the left-hand side of (3), that is, c(P ) = s. Meanwhile, if P is
contained in V (Q
Q
) ∩ ∂Q
Q
and t edges of Q
Q
extend from P , there exist
t − 1 faces containing P . It follows that c(P ) = t − 1. Consequently, we
obtain
k
∑
j=1
(aj +bj) = 2l(Int(Q
Q
))+l(∂Q
Q
\V (Q
Q
))+
s0
∑
s=3
(s−2)ms +
t0
∑
t=2
(t−1)nt,
270 The lower bound for the volume of a polytope
where we define
ms = ♯{P ∈ V (Q
Q
) ∩ Int(Q
Q
) | there exist s edges of Q
Q
extending from P},
nt = ♯{P ∈V (Q
Q
)∩∂Q
Q
| there exist t edges of Q
Q
extending from P}.
Next, to compute the value of k, we take a lattice point P0 /∈ Conv(Q
Q
)
such that an integral polytope Q0 = Conv(Q
Q
∪ {P0}) satisfies V (Q0) =
V (Q
Q
) ∪ {P0} (see Fig. 1).
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❳❳❳❳❳❳❳❳❳❳❳❳
❛❛❛❛❛❛❛❛❛❛❛❛ ✭✭✭✭✭✭✭✭✭✭✭✭◗◗
◗◗
◗◗
◗◗
❇❇❇❇❇❇❇❇
❆❆❆❆❆❆❆❆
✟✟✟✟✟✟✟✟❊
❊❊
❊
❊❊
❊
❊❊
❊
❊❊ ☞
☞☞
☞☞
☞☞
☞
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✥✥✥✥✥✥✥✥✥✥
✥✥✥✥✥✥✥✥✥✥
✥✥✥✥✥✥✥✥✥✥
✥✥✥✥✥✥✥✥✥✥P0
Q0
Figure 1.
Then the number of vertices, edges and faces of Q0 are
∑s0
s=3 ms +
∑t0
t=2 nt + 1,
(
∑s0
s=3 sms +
∑t0
t=2 tnt
)
/2 +
∑t0
t=2 nt and k +
∑t0
t=2 nt, respec-
tively. Hence, by Euler’s polyhedron formula, we have
s0
∑
s=3
ms +
t0
∑
t=2
nt + 1 −
s0
∑
s=3
sms +
t0
∑
t=2
tnt
2
−
t0
∑
t=2
nt + k +
t0
∑
t=2
nt = 2,
which implies that k =
(
∑s0
s=3(s − 2)ms +
∑t0
t=2(t − 2)nt
)
/2 + 1. As a
consequence,
µ(Q) − µ(QQ) =
1
6
(
l(∂Q
Q
\ V (Q
Q
)) +
t0
∑
t=2
nt
)
−
2
3
+
1
6
k
∑
j=1
(hj − 1)(aj + bj − 2)
=
1
6
l(∂Q
Q
) −
2
3
+
1
6
k
∑
j=1
(hj − 1)(aj + bj − 2).
R. Kawaguchi 271
By noting that l(∂Q
Q
) > 3, aj > 3 and bj > 0, we obtain the following
corollary.
Corollary 2.4. Let Q, Q and Fj be as in Proposition 2.3. If there exists
a face Fj0
of QQ such that h(Fj0
, Q) > 2, then µ(Q) > µ(QQ).
Let us show the main result. Since the proof is relatively long, we
divide it into two parts.
Proof of the inequality in Theorem 1.4. Let T be a section of P as in
Lemma 2.2. We take a lattice point P0 ∈ V (P) \ T and put P1 = PP0 .
By carrying out such operation repeatedly, we construct a sequence of
integral convex polytopes
P = P0 → P1 → P2 → · · · , (4)
where Pi ∈ V (Pi) \ T and Pi+1 = PPi
i . It is sufficient to show that, by
going through a suitable process, we can find Pn such that µ(P0) > µ(Pn),
l(Pn) = l(T ) + 2 and T is not a face of Pn (see Fig. 2).
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❵❵❵❵❵❵❵❵❵❵❵❵❵❵❵❵❵❵❵❵❵❵❵❵❵❵❵❵❵❵❵❵❵❵❵❵❵❵❵❵❵❵❵❵❵✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏
✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏✏
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♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣
♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣
❤❤❤❤❤
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❤❤❤❤❤ ❤❤❤❤❤
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◗❵❵❵❵❵❵❵❵❵❵❵❵✏✏✏✏✏✏✏✏✏✏✏✏
✚
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✚
✚✚
✚
✚✚T ◗
◗
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◗
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◗❵❵❵❵❵❵❵❵❵❵❵❵✏✏✏✏✏✏✏✏✏✏✏✏
✚
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❛❛❛❛❛
❛❛❛❛❛
❛❛❛❛❛
❛❛❛❛❛❝❝❝❝❝❝❝❝ ✪
✪
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✪
✪
✪
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✪
✪
✪
✪
✪
✪
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PP
PP
PPP
PP
PP
PPP
PP
PP
PPP
PP
PP
PPP❅
❅
❅
❅❅
❅
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❅
❅
❅
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❅
❅
❅
❅❅
Pn
Figure 2.
Indeed, for such a Pn, it follows from Theorem 1.1 that
µ(Pn) = vol(Pn) −
l(Pn) + l(Int(Pn)) − 4
3
>
2
3
vol(T ) −
l(T ) + l(Int(T )) − 2
3
=
2
3
·
l(T ) + l(Int(T )) − 2
2
−
l(T ) + l(Int(T )) − 2
3
= 0.
To verify the existence of Pn, it is sufficient to prove the following
claim:
Claim A. Let i be a nonnegative integer. If l(Pi) > l(T )+3 and T is not
a face of Pi, then we can construct Pi0
(i0 > i) such that µ(Pi) > µ(Pi0
)
and T is not a face of Pi0
.
272 The lower bound for the volume of a polytope
We define A = {Q ∈ V (Pi) \ T | T is not a face of PQ
i }. If there
exists a point Q ∈ A such that l(∂Pi
Q
) > 4, by putting Pi = Q, we obtain
the inequality µ(Pi) > µ(Pi+1) by Proposition 2.3. Hence Claim A is true
in this case. We thus assume that
l(∂Pi
Q
) = 3 for any Q ∈ A. (5)
We take a point Q0 ∈ A. Note that the inequality µ(Pi) > µ(PQ0
i )−1/6
follows from Proposition 2.3. We denote by Q1, Q2 and Q3 the vertices
of a triangle ∂Pi
Q0 , and put vj = Qj − Q0 for j = 1, 2, 3. We define
εj = max{ε ∈ N | Q0 + εvj ∈ Pi} and Q′
j = Q0 + εjvj for j = 1, 2, 3.
(i) We first consider the case where l(Conv({Q0, Q1, Q2, Q3})) > 5.
We put t = ♯({Q1, Q2, Q3} ∩ T ). If t = 3, then T is a triangle
Conv({Q1, Q2, Q3}) whose border has no lattice points except for three
vertices. This contradicts the property of T of having a smooth vertex
and interior lattice points. Hence we have t 6 2.
(i)–(a) If t 6 1, we can assume Q1, Q2 /∈ T and ε1 6 ε2.
In the case where ε1 > 2, we put Pi = Q0, Pi+1 = Q1 and
Pi+2 = Q2. Then ∂Pi+1
Pi+1 contains Q1 + v1, Q2, Q3 and at least
one lattice point in Conv({Q0, Q1, Q2, Q3}) \ {Q0, . . . , Q3}. It follows
from Proposition 2.3 that µ(Pi+1) > µ(Pi+2). Similarly, since ∂Pi+2
Pi+2
contains Q1 + v1, Q1 + v2, Q2 + v2, Q3 and at least one lattice point in
Conv({Q0, Q1, Q2, Q3})\{Q0, . . . , Q3}, we have µ(Pi+2) > µ(Pi+3)+1/6.
In sum, Claim A is true by
µ(Pi) > µ(Pi+1) −
1
6
> µ(Pi+2) −
1
6
> µ(Pi+3).
We next consider the case where ε1 = 1. Since Q1 ∈ A, we have
l(∂Pi
Q1) = 3 by (5), and more concretely, Q′
1(= Q1) has three adjacent
lattice points Q′
1 − v1(= Q0), Q0 + αv1 + βv2 and Q0 + γv1 + δv3 in
E(Pi), where α, γ > 0 and β, δ > 1. We denote by F a face of Pi
Q1
containing two points Q0 and Q0 + αv1 + βv2. By an easy computation,
we obtain h(F, Q1) > β. If β > 2, we can finish the proof by putting
Pi = Q1. Indeed, in this case, the inequality µ(Pi) > µ(Pi+1) holds by
Corollary 2.4. Similar arguments can be carried out for the case where
δ > 2. Let us consider the case where β = δ = 1. We can assume α > γ
without loss of generality, and have α + γ > 1 by the existence of T .
If α + γ > 2, we put Pi = Q0 and Pi+1 = Q1. Then, since ∂Pi+1
Pi+1
R. Kawaguchi 273
contains Q2 + kv1 (k = 0, . . . , α), Q3 + kv1 (k = 0, . . . , γ) and at least
one lattice point in Conv({Q0, Q1, Q2, Q3}) \ {Q0, . . . , Q3}, we obtain
µ(Pi+1) > µ(Pi+2) + (α + γ − 1)/6. Hence
µ(Pi) > µ(Pi+1) −
1
6
> µ(Pi+2) +
α + γ − 2
6
> µ(Pi+2).
Let us consider the remaining case where α = 1 and γ = 0. Note that
Pi has a face containing Q1, Q3 and Q1 + v2 (see Fig. 3).
✲✄
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✦✦
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✦✦✯
Q0 Q1
Q2
Q3
v1
v2
v3
✄
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✦✦
q
Q0 Q1
Q2
Q3 Pi
Figure 3.
We define ζj = max{ζ ∈ Z>0 | Qj + ζv2 ∈ Pi} for j = 1, 2, 3.
Considering the properties of T , at least one of Q1 + ζ1v2, Q2 + ζ2v2, and
Q3 + ζ3v2 is not on T . Then, by (5), such a point has just three adjacent
lattice points in E(Pi). This implies that V (Pi) = {Q0, Q1, Q3, Q1 +
ζ1v2, Q2 + ζ2v2, Q3 + ζ3v2}, which contradicts the existence of T .
(i)–(b) If t = 2, we can assume that Q1 /∈ T and Q2, Q3 ∈ T . By the
properties of T , we see that Q0 + εv1 is not on T for 0 6 ε 6 ε1. As
we saw in the case (i)-(a), Q′
1 has three adjacent lattice points Q′
1 − v1,
Q0 + αv1 + βv2 and Q0 + γv1 + δv3 in E(Pi), and the proof is finished by
putting Pi = Q′
1 in the case where β > 2 or δ > 2. We assume β = δ = 1.
Since T has interior lattice points, we see that at least one of Q0+αv1+v2
and Q0 + γv1 + v3 is not on T . Then this case is equivalent to the case
(i)-(a) by regarding Q′
1 as Q0.
(ii) We next consider the case where l(Conv({Q0, Q1, Q2, Q3})) = 4.
If h(Conv({Q1, Q2, Q3}), Q0) > 2, we can finish the proof by putting
Pi = Q0. Hence we can assume that Q0 is a smooth vertex, that is,
Q0 = O, Q1 = (1, 0, 0), Q2 = (0, 1, 0) and Q3 = (0, 0, 1). Moreover, we
assume that every vertex in A is smooth in order to avoid the duplication
with the case (i). We denote by Lj the segment Q0Q′
j (j = 1, 2, 3), and
put u = ♯{Lj | Lj ∩ T 6= ∅, j = 1, 2, 3}.
(ii)–(a) In the case where u 6 1, we can assume that L1 ∩ T = ∅
and L2 ∩ T = ∅. Since Q′
1 is smooth, it has three adjacent lattice points
(ε1 − 1, 0, 0), (α, 1, 0) and (γ, 0, 1) in E(Pi). If we put Pi+ε = (ε, 0, 0)
274 The lower bound for the volume of a polytope
for ε = 0, . . . , ε1, since ∂Pi+ε1
Pi+ε1 contains lattice points (k, 1, 0) with
k = 0, . . . , α and (l, 0, 1) with l = 0, . . . , γ, it follows that
µ(Pi) > µ(Pi+1)−
1
6
> · · ·> µ(Pi+ε1
)−
ε1
6
> µ(Pi+ε1+1)+
α + γ − ε1 − 2
6
.
If α + γ > ε1 + 2, the proof is finished. We next consider the vertex
Q′
2 and its three adjacent lattice points (0, ε2 − 1, 0), (1, β, 0) and (0, δ, 1)
in E(Pi). Similarly to the case of Q′
1, we can finish the proof in the
case where β + δ > ε2 + 2. Hence we assume that α + γ 6 ε1 + 1 and
β + δ 6 ε2 + 1. Let (x0, y0, z0) be a lattice point in Int(Pi). Since Pi has a
face containing three points (ε1, 0, 0), (α, 1, 0) and (γ, 0, 1) (resp. (0, ε2, 0),
(1, β, 0) and (0, δ, 1)), we have
x0 + (ε1 − α)y0 + (ε1 − γ)z0 − ε1 < 0
(resp. (ε2 − β)x0 + y0 + (ε2 − δ)z0 − ε2 < 0).
By noting x0, y0, z0 > 1 and α, γ, β, δ > 0, we obtain (α, γ) = (ε1 + 1, 0)
or (0, ε1 + 1) and (β, δ) = (ε2 + 1, 0) or (0, ε2 + 1). Clearly, γ = δ = 0
gives a contradiction. Besides, considering the shape of Pi, if either α or
β is zero, then the other one also must be zero and ε1 = ε2 = 1. In sum,
we have (α, γ) = (β, δ) = (0, 2). Then, putting Pi = Q0 and Pi+j = Qj
for j = 1, 2, we have
l(∂Pi
Pi) = l(Conv({(1, 0, 0), (0, 1, 0), (0, 0, 1)})) = 3,
l(∂Pi+1
Pi+1) = l(Conv({(2, 0, 1), (0, 1, 0), (0, 0, 1)})) = 4,
l(∂Pi+2
Pi+2) = l(Conv({(2, 0, 1), (0, 2, 1), (0, 0, 1)})) = 6.
It follows from Proposition 2.3 that µ(Pi) = µ(Pi+1)−1/6 = µ(Pi+2)−
1/6 = µ(Pi+3) + 1/6.
(ii)–(b) If u = 2, we can assume that L1 ∩ T = ∅, L2 ∩ T 6= ∅ and
L3 ∩ T 6= ∅. As we saw in the case (ii)-(a), it is sufficient to consider the
case where Q′
1 has three adjacent lattice points (ε1 − 1, 0, 0), (α, 1, 0) and
(γ, 0, 1) with (α, γ) = (ε1 + 1, 0) or (0, ε1 + 1). Here we consider only the
former case (α, γ) = (ε1 + 1, 0). The latter case can be shown in a similar
way. First, we remark that ε3 = 1 holds by the condition γ = 0. This
means that (0, 0, 1) is on T . Denote by L4 the line passing through Q′
1
and (α, 1, 0). Since T has interior lattice points, L4 does not contain a
lattice point on T . Then, by regarding Q′
1 as Q0, this case can be reduced
to the case (ii)-(a).
R. Kawaguchi 275
(ii)–(c) Assume u = 3. In this case, we see that Qj /∈ T
for j = 1, 2, 3 since T has interior lattice points. It follows that
(2, 0, 0), (0, 2, 0), (0, 0, 2) ∈ Pi. If we put Pi = Q0 and Pi+j = Qj for
j = 1, 2, 3, then
l(∂Pi
Pi) = l(Conv({(1, 0, 0), (0, 1, 0), (0, 0, 1)})) = 3,
l(∂Pi+1
Pi+1) = l(Conv({(2, 0, 0), (0, 1, 0), (0, 0, 1)})) = 3,
l(∂Pi+2
Pi+2) = l(Conv({(2, 0, 0), (0, 2, 0), (0, 0, 1)})) = 4,
l(∂Pi+3
Pi+3) = l(Conv({(2, 0, 0), (0, 2, 0), (0, 0, 2)})) = 6.
Hence we have
µ(Pi) > µ(Pi+1) −
1
6
> µ(Pi+2) −
2
3
> µ(Pi+3) −
2
3
> µ(Pi+4).
Since T has interior lattice points, at least one of (2, 0, 0), (0, 2, 0) and
(0, 0, 2) is not on T , that is, T is not a face of Pi+4.
In order to show the latter part of Theorem 1.4, we require results
in the theory of toric varieties and the classification theory of polarized
varieties. Hence, in the proof below, we take in advance the contents in
the next section although not in the proper order. See Section 3 for precise
definitions and notations.
Proof of the equivalency in Theorem 1.4. The classification of toric Fano
threefolds has been completed, and they are classified into eighteen types
(cf. [2, 17]). For each type X of toric Fano threefolds, the polytope �−KX
associated to the anti-canonical bundle has just one interior lattice point.
Moreover, we can obtain
vol(�−KX
) =
l(�−KX
)
3
− 1
by steady calculations. We list several examples of them for readers’
exercise.
X �−KX
P
3 the fourth dilation of a unit three-simplex
P
1 × P
1 × P
1 the twice dilation of a unit cube
P
2 × P
1 Conv({(0, 0, ±1), (3, 0, ±1), (0, 3, ±1)})
Σ1 × P
1 Conv({(0, 0, ±1), (3, 0, ±1), (1, 2, ±1), (0, 2, ±1)})
...
...
(6)
276 The lower bound for the volume of a polytope
Let us show the sufficiency. We first consider the case where
l(Int(P))=1 and vol(P) = l(P)/3 − 1. Let L be an ample line bundle
on a three-dimensional toric variety X whose associated polytope �L
coincides with P. Then our assumptions are equivalent to two equalities
h0(X, L + KX) = 1 and L3 = 2h0(X, L) − 6. By noting Lemma 3.7,
the sectional genus and the ∆-genus of the polarized variety (X, L) is
g(X, L) = h0(X, L) − 2 and ∆(X, L) = h0(X, L) − 3, respectively. On the
other hand, since ⌊(L3 − 1)/(L3 − ∆(X, L) − 1)⌋ = 2, the above sectional
genus coincides with the upper bound in Theorem 3.3. Namely, (X, L) is
a Castelnuovo variety in this case. Since (X, L) is a Mukai variety by the
remark after Theorem 3.3, we can conclude that X is a Fano variety and
L = −KX .
In the remaining part, we prove the inequality vol(P) > (l(P) +
l(Int(P)) − 4)/3 under the assumption l(Int(P)) > 2. Recall the sequence
of integral convex polytopes (4) and Claim A in the proof of Theorem 1.4.
Then it is sufficient to show that, by going through a suitable process,
we can construct Pi0
such that µ(P) > µ(Pi0
) and T is not a face of Pi0
.
We place P in R
3 so that four points O, Q1 = (1, 0, 0), Q2 = (0, 1, 0) and
Q3 = (0, 0, 1) are contained in E(P), and define
ε1 = max{ε ∈ N | (ε, 0, 0) ∈ P},
ε2 = max{ε ∈ N | (0, ε, 0) ∈ P},
ε3 = max{ε ∈ N | (0, 0, ε) ∈ P},
Q′
1 = (ε1, 0, 0), Q′
2 = (0, ε2, 0) and Q′
3 = (0, 0, ε3). Then, by the smooth-
ness of P, we see that Q′
1 has three adjacent lattice points (ε1 − 1, 0, 0),
(α1, 1, 0) and (α2, 0, 1) in E(P). Similarly, Q′
2 (resp. Q′
3) has three adja-
cent lattice points (0, ε2 − 1, 0), (1, β1, 0) and (0, β2, 1) (resp. (0, 0, ε3 − 1),
(1, 0, γ1) and (0, 1, γ2)) in E(P). In the case where α1 + α2 > ε1 + 3,
we can take T so that it does not contain points on the x-axis by us-
ing a similar method to that in the proof of Lemma 2.2. If we put
Pi = (i, 0, 0) for i = 0, . . . , ε1, then ∂Pi
Pi is a triangle with vertices
(i + 1, 0, 0), Q2 and Q3 for i = 0, . . . , ε1 − 1, and ∂Pε1
Pε1 is a trapezoid
Conv({Q2, Q3, (α1, 1, 0), (α2, 0, 1)}). We thus obtain
µ(P0) > µ(P1) −
1
6
> · · · > µ(Pε1
) −
ε1
6
> µ(Pε1+1) +
α1 + α2 − ε1 − 2
6
> µ(Pε1+1)
by Proposition 2.3. Also in the cases where β1 + β2 > ε2 + 3 or γ1 + γ2 >
ε3 + 3, we can finish the proof in essentially the same way.
R. Kawaguchi 277
We assume henceforth that α1 + α2 6 ε1 + 2, β1 + β2 6 ε2 + 2 and
γ1 + γ2 6 ε3 + 2. Moreover, without loss of generality, we can assume that
OQ′
1 is the shortest edge of P and α1 > α2. Since P has a face containing
three points Q′
1, (α1, 1, 0) and (α2, 0, 1), the inclusion
Int(P) ⊂ {(x, y, z) | x, y, z > 1, x + (ε1 − α1)(y − 1)
+ (ε1 − α2)(z − 1) + ε1 − α1 − α2 + 1 6 0}
(7)
is derived from the inequality x+(ε1−α1)y+(ε1−α2)z−ε1 < 0. Similarly,
by considering the vertices Q′
2 and Q′
3, we obtain
Int(P) ⊂ {(x, y, z) | x, y, z > 1, (ε2 − β1)(x − 1) + y
+ (ε2 − β2)(z − 1) + ε2 − β1 − β2 + 1 6 0},
(8)
Int(P) ⊂ {(x, y, z) | x, y, z > 1, (ε3−γ1)(x−1)+(ε3−γ2)(y−1)
+ z + ε3 − γ1 − γ2 + 1 6 0}.
(9)
(i) Assume that α1 6 ε1, and let (x0, y0, z0) be a lattice point in
Int(P). By noting α1 > α2 and α1 + α2 6 ε1 + 2, we have x0 = 1 and
α1 + α2 = ε1 + 2 by (7). If α2 < ε1, we see that z0 = 1 by (7) and y0 = 1
by (8), which contradicts the assumption l(Int(P)) > 2. We thus have
α2 > ε1, and similarly β2 > ε2 and γ2 > ε3. Since ε1 = α1 = α2 = 2
in this case, ε2, ε3 6 2 follows from the shortestness of the edge OQ′
1.
Moreover, it follows from (2, 1, 0), (2, 0, 1) ∈ P that β1 > ε2 and γ1 > ε3.
As a consequence, we have (ε2, β1, β2) = (ε3, γ1, γ2) = (1, 1, 2). By the
shortestness of the edge OQ′
1 again, the faces P ∩ Hx and P ∩ Hx−2 are
one of the four types of polygons as in Fig. 4.
✲✲✲
✻✻✻
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��������
y
z
G1
✲✲✲
✻✻✻
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y
z
G2
✲✲✲
✻✻✻
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�
�
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�
�
��
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y
z
G3
✲✲✲
✻✻✻
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y
z
G4
Figure 4.
Then, considering the smoothness of P and the assumption l(Int(P)) >
2, there exist only three possibilities (P ∩ Hx, P ∩ Hx−1, P ∩ Hx−2) =
(G1, G4, G4), (G4, G4, G1), (G4, G4, G4). In the first two cases, we have
vol(P) = 9, l(P) = 27 and l(Int(P)) = 2. On the other hand, in the last
case, vol(P) = 10, l(P) = 30 and l(Int(P)) = 2. Hence the inequality
vol(P) > (l(P) + l(Int(P)) − 4)/3 holds in each case.
278 The lower bound for the volume of a polytope
(ii) Suppose that ε1 + 1 6 α1 6 ε1 + 2 and β1, β2 6 ε2. Let (x0, y0, z0)
be a lattice point in Int(P). We have y0 = 1 and β1 + β2 = ε2 + 2 by (8).
Then, since ε1 − α2 > α1 − 2 > 0, we have x0 = 1 and α1 + α2 = ε1 + 2
by (7). By the assumption l(Int(P)) > 2, there must be an interior
lattice point such that z0 > 2. It follows that α2 = ε1 and β2 = ε2.
These facts, together with the shortestness of OQ′
1, immediately give that
ε1 = ε2 = α2 = β2 = 1 and α1 = β1 = 2. By using the shortestness of
OQ′
1 again, we see that P ∩ Hx and P ∩ Hy are unit squares, which yields
a contradiction l(Int(P)) = 0.
(iii) Suppose that ε1+1 6 α1 6 ε1+2 and β2 > ε2+1. Note that α1 > 2
and β1 6 1 in this case. Then β1 must be one since (2, 1, 0) ∈ P. Hence
P ∩ Hz is a trapezoid Conv({O, Q′
1, (α1, 1, 0), Q2}), which contradicts the
shortestness of OQ′
1.
(iv) Suppose that ε1 + 1 6 α1 6 ε1 + 2 and β2 = 0. In this case, ε2
must be one by the smoothness of the vertex Q3. We thus have ε3 = 1
and γ2 = 0, which imply γ1 > 2 (namely, (1, 0, 2) ∈ P) by (9). By noting
(2, 0, 1) /∈ P, we see that P∩Hy is a trapezoid Conv({O, Q1, (1, 0, γ1), Q3}),
which contradicts the shortestness of OQ′
1.
(v) We finally consider the remaining case where ε1 + 1 6 α1 6 ε1 + 2,
β1 = ε2 + 1 and β2 = 1. If ε1 = 1, by the shortestness of OQ′
1, P ∩ Hx
is a unit square, and P ∩ Hy is a unit triangle or a unit square. This
contradicts the assumption l(Int(P)) > 2. We thus assume ε1 > 2. Note
that α1 = ε1 + 1 and α2 = 1 in this case. Hence, if (x0, y0, z0) is a lattice
point in Int(P), x0 = y0 and z0 = 1 follow from (7) and (8). We define
ε4 = max{ε ∈ N | (ε, ε2 + ε, 0) ∈ P}, and put Q4 = (ε4, ε2 + ε4, 0). By
the smoothness of P, the vertex Q4 has three adjacent lattice points
(ε4 − 1, ε2 + ε4 − 1, 0), (δ1, δ1 + ε2 − 1, 0) and (δ2, δ2 + 1, 1) in E(P) (see
Fig. 5).
✲✲
✻✻
�
�
��
�
�
��
�
�
��
�
�
��
✟✟
✟
✟✟
✟
✟✟
✟
✟✟
✟
✟✟
✟
t
t
t
x
y
O
Q′
2
Q4
(δ1, δ1 + ε2 − 1, 0)
P ∩ Hz
✲✲
✻✻
✑✑✑✑
✸✸
s
s
s
s
s
Q′
2 Q4
(δ1, δ1 + ε2 − 1, 0)
(δ2, δ2 + 1, 1)
(0, 1, 1)
x
y
z
Figure 5.
R. Kawaguchi 279
If δ1 + δ2 > ε4 + 3, we put Pi = (i, ε2 + i, 0) for i = 0, . . . , ε4. Then
∂Pi
Pi is a triangle with vertices (i + 1, ε2 + i + 1, 0), (0, ε2 − 1, 0) and
(0, 1, 1) for i = 0, . . . , ε4 − 1, and ∂Pε4
Pε4 is a trapezoid
Conv({(0, ε2 − 1, 0), (δ1, δ1 + ε2 − 1, 0), (0, 1, 1), (δ2, δ2 + 1, 1)}).
The proof is finished since
µ(P0) > µ(P1) −
1
6
> · · · > µ(Pε4
) −
ε4
6
> µ(Pε4+1) +
δ1 + δ2 − ε4 − 2
6
> µ(Pε1+1)
by Proposition 2.3. Finally, let us show that the case where δ1 + δ2 6
ε4 + 2 does not occur. Since P has a face containing three points Q4,
(δ1, δ1 + ε2 − 1, 0) and (δ2, δ2 + 1, 1), the inclusion
Int(P) ⊂ {(x, y, z) | x, y, z > 1, (ε4 − δ1 + 1)(x − 1) − (ε4 − δ1)(y − 1)
+ (δ1ε2 − ε2ε4 − δ1 − δ2 + 2ε4)(z − 1)
− δ1 − δ2 + ε4 + 2 6 0}
holds. As we have already mentioned, any interior lattice point of P can
be written as (x0, x0, 1). This contradicts the above inclusion and the
assumption l(Int(P)) > 2.
3. Application
In this section, we apply our result to the computation of the sectional
genus of a polarized toric variety. For an n-dimensional (smooth) complex
projective variety X and an ample line bundle L on X, the pair (X, L) is
called a (smooth) polarized variety. We remark that, in the case where L
is ample, the associated polytope �L is smooth if and only if X is smooth.
Let us review the classification theory of polarized varieties before getting
to the main subject. We first recall the well-known upper bound for the
geometric genus of a smooth curve.
Theorem 3.1 (Castelnuovo’s bound, [1]). Let C be a smooth curve of
genus g. Assume that C admits a birational map onto a nondegenerate
curve of degree d in P
r. Then
g 6
1
2
a(a − 1)(r − 1) + a(d − a(r − 1) − 1),
where a = ⌊(d − 1)/(r − 1)⌋.
280 The lower bound for the volume of a polytope
A smooth curve is said to be extremal if its genus is equal to Casteln-
uovo’s bound, which was studied in [4]. As a higher dimensional extension,
Fujita established various invariants for polarized varieties and proved a
similar inequality (Theorem 3.3).
Definition 3.2. For an n-dimensional smooth polarized variety (X, L),
we define the sectional genus and the ∆-genus by
g(X, L) =
1
2
Ln−1.((n − 1)L + KX) + 1,
∆(X, L) = Ln + n − h0(X, L).
Theorem 3.3 (cf. [5]). Let L be a line bundle on an n-dimensional
smooth projective variety X. If |L| has no base points and the associated
morphism Φ|L| is birational on its image, then
g(X, L) 6 a∆(X, L) −
1
2
a(a − 1)(Ln − ∆(X, L) − 1),
where a = ⌊(Ln − 1)/(Ln − ∆(X, L) − 1)⌋.
We call (X, L) a Castelnuovo variety if its sectional genus achieves the
maximum of the above upper bound. Castelnuovo varieties can be roughly
classified according to the relation between Ln and 2∆(X, L). First, the
case where Ln < 2∆(X, L) has been classified in [5]. If Ln = 2∆(X, L),
then (X, L) is a Mukai variety (i.e., KX ∈ |(2 − n)L|), which has been
classified in [11]. On the other hand, the case where Ln > 2∆(X, L) still
has many unknown aspects. Two- or three-dimensional polarized toric
varieties, which we consider below, contain examples of this case.
In the two-dimensional case, we do not need to use the results in this
paper, but only the Riemann-Roch theorem. To see this, let us introduce
the notion of a ladder.
Definition 3.4. Let (X, L) be an n-dimensional polarized variety, and
put X0 = X and L0 = L. A sequence X0 ⊃ X1 ⊃ · · · ⊃ Xn−1 of (smooth)
subvarieties of X is called a (smooth) ladder of (X, L) if Xi ∈ |Li−1| for
each i > 1, where we put Li = L|Xi
.
Theorem 3.5 (cf. [5]). Let (X, L) be an n-dimensional polarized variety
having a ladder. If Ln > 2∆(X, L), then L is very ample and g(X, L) =
∆(X, L).
R. Kawaguchi 281
If X is smooth and L is generated by global sections, by virtue of
Bertini’s theorem, we obtain a smooth ladder of (X, L) by cutting Xi
by a general member of |Li|. Since any ample line bundle on a compact
toric variety is generated by global sections, a polarized toric variety
always has a smooth ladder. Using these results, let us now consider the
two-dimensional polarized toric varieties.
Theorem 3.6. For a smooth compact toric surface X and an ample line
bundle L on X, the polarized variety (X, L) is a Castelnuovo variety with
L2 > 2∆(X, L) + 2 unless L is a line in P
2.
Proof. By the general theory of toric varieties, L is very ample, and we have
pa(X) = 0, h0(X, L) > 3 and h1(X, L) = h2(X, L) = 0. Hence we obtain
L2 = 2h0(X, L) + L.KX − 2 by the Riemann-Roch theorem. On the other
hand, since −L.KX is equal to l(∂�L), we have the inequality −L.KX > 3,
where the equality holds if and only if (X, L) ≃ (P2, OP2(1)). Consequently,
we have L2 6 2h0(X, L)−6, which implies that L2 > 2∆(X, L)+2. Then,
since g(X, L) = ∆(X, L) by Theorem 3.5, we can conclude that (X, L) is
a Castelnuovo variety.
Next, in order to investigate the three-dimensional case, we compute
the value of L2.KX . This computation also can be reduced to a matter
of the number of lattice points.
Lemma 3.7. For a three-dimensional smooth polarized toric variety
(X, L), the equality L2.KX = 2(h0(X, L + KX) − h0(X, L) + 2) holds.
Proof. We denote by D1, . . . , Dd the TN -invariant divisors of X, and by
Fi the face of �L corresponding to Di. It is known that KX ∼ −
∑d
i=1 Di
and L2.Di is equal to twice of the area of Fi. Hence the statement of the
lemma can be rewritten as
d
∑
i=1
vol(Fi) = −l(Int(�L)) + l(�L) − 2 = l(∂�L) − 2.
Let us compute the left-hand side by using Theorem 1.1 and Euler’s
polyhedron formula. Denote by v and e the number of vertices and edges
of �L, respectively. It follows from the smoothness of �L that every vertex
of �L has three edges. Hence we have 3v = 2e and
d
∑
i=1
vol(Fi) =
1
2
d
∑
i=1
(l(Fi) + l(Int(Fi)) − 2) =
1
2
(2l(∂�L) + v) − d
= l(∂�L) − v + e − d = l(∂�L) − 2.
282 The lower bound for the volume of a polytope
Theorem 3.8. Let X be a three-dimensional smooth compact toric variety,
and L be an ample line bundle on X. Assume that (X, L) 6≃ (P3, OP3(1)).
Then a polarized variety (X, L) is a Castelnuovo variety if and only if
(i) X is a Fano variety and L ∼ −KX , or
(ii) h0(X, L + KX) = 0 and h0(X, 2L + KX) 6 h0(X, L) − 4.
We provide several additional explanations. The former case has been
classified into eighteen types in [2] and [17]. Besides, L3 = 2∆(X, L)
holds in this case. On the other hand, L3 > 2∆(X, L) holds in the
case (ii), and it is known that three-dimensional polarized toric varieties
with h0(X, L + KX) = 0 (not necessarily assume the latter inequality)
can be classified into five types (see [13]). We will see further details of
this classification after the proof. Incidentally, more generally, Fukuma
classified n-dimensional polarized varieties (not necessarily toric) with
h0(X, (n − 2)L + KX) = 0 in [6].
Proof of Theorem 3.8. If we rewrite the inequality in Theorem 3.1 by
using Lemma 3.7, we see that (X, L) is a Castelnuovo variety if and
only if
(a2 + a − 2)h0(X, L) = 2(2a2 + a − 3 + (a − 1)L3 − h0(X, L + KX)).
(10)
On the other hand, we have h0(X, L) 6 (L3 −1)/a+4 by the definition
of a. By combining this inequality with (10), we see that
2ah0(X, L + KX) > (a − 1)(a − 2)(L3 − 1) (11)
holds if (X, L) is a Castelnuovo variety.
(i) We first consider the case where h0(X, L + KX) > 1. In order to
prove the sufficiency, we assume that (X, L) is a Castelnuovo variety. Note
that a > 2 by (10). By Corollary 1.6 and Lemma 3.7, we have
h0(X, L+KX) 6 h0(X, L+KX)+
1
4
(L3−2h0(X, L)−2h0(X, L+KX)+8)
=
1
4
L3 −
1
2
h0(X, L) +
1
2
(
1
2
L2.KX + h0(X, L) − 2
)
+ 2
=
1
4
L3 +
1
4
L2.KX + 1 6
1
4
L3,
where the last inequality follows from the fact that −L2.KX is twice the
sum of areas of faces of �L. Then the inequality (11) induces (2a2 − 7a +
R. Kawaguchi 283
4)L3 6 2(a − 1)(a − 2). If a > 3, we have vol(�L) = L3/6 6 2/3, which
clearly contradicts the assumption l(Int(�L)) > 1. Hence we obtain a = 2.
Then, since
h0(X, L + KX) = L3 − 2h0(X, L) + 7 > 2h0(X, L + KX) − 1
by (10) and Corollary 1.6, we have h0(X, L+KX) = 1 and L3 = 2∆(X, L).
Hence (X, L) is a Mukai variety, which means that (X, L) ≃ (X, −KX)
is a Fano variety. Such varieties, so-called toric Fano three-folds, have
been classified into eighteen types in [2] and [17] independently (see (6)
in Section 2). Hence the necessity is checked by computing. For all types,
in practice, we can confirm that a = 2 and the equality (10) holds.
(ii) Assume that h0(X, L + KX) = 0 (equivalently, l(Int(�L)) = 0).
In this case, Theorem 1.2 gives
12vol(�L) = l(�2L) + l(Int(�2L)) − 2l(�L)
= l(∂�2L) + 2l(Int(�2L)) − 2l(�L)
= 4l(∂�L) + 2l(Int(�2L)) − 2l(�L) − 6
= 2l(�L) + 2l(Int(�2L)) − 6,
which implies that L3 = h0(X, L) + h0(X, 2L + KX) − 3. Therefore, the
inequality in the statement is equivalent to L3 6 2h0(X, L)−7. If (X, L) is
a Castelnuovo variety, we have a 6 2 by (11). In the case where a = 1, it is
clear that L3 < 2h0(X, L) − 7 by the definition of a. On the other hand, if
a = 2, the condition (10) is equivalent to the equality L3 = 2h0(X, L) − 7.
Conversely, if L3 6 2h0(X, L) − 7, we have
a =
{
1 (L3 < 2h0(X, L) − 7),
2 (L3 = 2h0(X, L) − 7)
by definition. In either case, we can easily check that (X, L) satisfies (10).
By virtue of [13, Proposition 2.3], we can see the detailed structure of
(X, L) in the case (ii) in Theorem 3.8. If h0(X, L + KX) = 0, then X is
one of the following five types.
(a) a toric P
1-bundle over a smooth toric surface.
(b) (X, L) ≃ (P3, OP3(k)) (k = 1, 2, 3).
(c) a toric P
2-bundle P(OP1(a) ⊕ OP1(b) ⊕ OP1(c)) over P
1.
284 The lower bound for the volume of a polytope
(d) a blow-up of P3 at TN -invariant i points (i = 1, 2, 3, 4).
(e) a blow-up of P(OP1(a) ⊕ OP1(b) ⊕ OP1(c)) at TN -invariant i points
(i = 1, 2).
In each case, the polytope �L is as follows. For simplicity, we put
m1 = (1, 0, 0), m2 = (0, 1, 0) and m3 = (0, 0, 1), and assume a > b > c
in the cases (c) and (e).
(a) Conv(F0∪F1), where F0 and F1 are parallel smooth faces of distance
one such that they define the same two-dimensional smooth fan.
(b) Conv({O, km1, km2, km3}).
(c) Conv({O, m1, m2, (1, 0, a), (0, 1, b), (0, 0, c)}) or
Conv({O, 2m1, 2m2, (2, 0, 2a − c), (0, 2, 2b − c), (0, 0, c)}).
(d) Conv({km1, km2, km3 | k = 1, 3}),
Conv({m1, m2, m3, 3m1, 3m2, 2m3, (1, 0, 2), (0, 1, 2)}),
Conv({m1, m2, m3, 3m1, 2m2, 2m3, (1, 2, 0), (0, 2, 1), (1, 0, 2),
(0, 1, 2)}) or Conv({km1, km2, km3, (2, 1, 0), (2, 0, 1), (1, 2, 0),
(0, 2, 1), (1, 0, 2), (0, 1, 2) | k = 1, 2}).
(e) a polytope obtained from Q by cutting of a unit three-simplex at
one of O, 2m1 and 2m2, where Q denotes the latter polytope in the
case (c),
a polytope obtained from the above one by cutting of a unit three-
simplex at one of (2, 0, 2a − c), (0, 2, 2b − c) and (0, 0, c).
By computing, we can check 6vol(�L) 6 2l(�L) − 8, that is,
(X, L) is a Castelnuovo variety in the latter four cases except for
Conv({O, m1, m2, m3}) (in which case L is a hyperplane in P
3). On the
other hand, in the case (a), the value of vol(�L) varies greatly depending
on the shapes of F0 and F1. See the following examples.
Example 3.9. Let (X, L) be a three-dimensional polarized toric variety
of type (a).
(a1) If Fi = Conv({(i, 0, 0), (i, 3, 0), (i, 3, 3), (i, 0, 3)}, then 6vol(�L) =
2l(�L) − 10.
(a2) If Fi = Conv({(i, 0, 0), (i, 2, 0), (i, 3, 1), (i, 3, 2), (i, 2, 3), (i, 1, 3),
(i, 0, 2)}), then 6vol(�L) = 2l(�L) − 7.
(a3) If Fi = Conv({(i, 1, 0), (i, 2, 0), (i, 3, 1), (i, 3, 2), (i, 2, 3), (i, 1, 3),
(i, 0, 2), (i, 0, 1)}), then 6vol(�L) = 2l(�L) − 6.
In the first two cases, (X, L) is a Castelnuovo variety, while the last
one is not.
R. Kawaguchi 285
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Contact information
R. Kawaguchi Department of Mathematics,
Nara Medical University, Kashihara,
Nara 634-8521, Japan
E-Mail(s): kawaguchi@naramed-u.ac.jp
Received by the editors: 13.04.2015
and in final form 27.07.2015.
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